How do you find the maximum or minimum of $y + 3 x^{2} = 9$ $y+3x^2=9$ ?

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Answer 1 · 2018-05-23T12:08:13

Maximum value of $y = 9$ $y=9$ at $x = 0$ $x=0$

$y + 3 x^{2} = 9 or y = - 3 x^{2} + 9 or y = - 3 {(x - 0)}^{2} + 9$ $y +3 x^2 =9 or y = -3 x^2 +9 or y= -3(x-0)^2+9$ .

This is equation of parabola opening downward since coefficient of

$x^{2}$ $x^2$ is negative. So minimum value will be $- \infty$ $- oo$ and

maximum value will be at vertex. Comparing with vertex form of

equation $f (x) = a {(x - h)}^{2} + k; (h, k)$ $f(x) = a(x-h)^2+k ; (h,k)$ being vertex we find

here $h=0 , k=9 :.$ Vertex is at $(0,9)$

Therefore the minimum value of function is $- oo$ and

maximum value of function is $y=9$ at $x=0$

graph{y+ 3x^2=9 [-22.5, 22.5, -11.25, 11.25]} [Ans]

How do you find the maximum or minimum of y+3x^2=9y+3x2=9y+3x^2=9?